Lazy Limits 9!

Relevant wiki: Taylor Series - Problem Solving

By Maclaurin's Series of sinx we use

sinx = x-x^3/6......................

Rest of the terms are insignificant to us

Putting in the above limit(numerator)

xsin(x-x^3/6)-(x-x^3/6)^2

Again use maclaurin's series of sinx and collect terms of x^6 in numerator

The coefficient will be 1/18

hence limit is 1/18 making the answer 1+18 = 19

Let $y=\sin x$ . Then as $x \to 0$ , $y \to 0$ . Therefore the limit advances to:

$\large \displaystyle \lim_{y \to 0} \frac{\arcsin \left(y\right)\sin \left(y\right)-y^2}{\left(\arcsin \left(x\right)\right)^6}$

Plugging in the Taylor's Expansions for $\sin y$ and $\arcsin y$ :

$\sin y = y-\frac{y^3}{3!}+\frac{y^5}{5!}-\frac{y^7}{7!}+...$

$\arcsin y = y+\frac{y^3}{6}+\frac{3y^5}{40}+...$

$\large \displaystyle \lim_{y \to 0} \frac{\left(y+\frac{y^3}{6}+\frac{3y^5}{40}+...\right)\left(y-\frac{y^3}{3!}+\frac{y^5}{5!}-...\right)-y^2}{\left(y+\frac{y^3}{6}+\frac{3y^5}{40}+...\right)}$

Now we arrange all $x^6$ terms in the numerator and the denominator: (For this limit to exist in the form $\frac{a}{b}$ all the terms with degree lesser than 6 must cancel out and terms with degree more than 6 must evaluate to 0 as $y \to 0$ )

$\large \displaystyle \lim_{y \to 0} \frac{\left(y\cdot \frac{y^5}{5!}-\frac{y^3}{6}\cdot \frac{y^3}{3!}+\frac{3y^5}{40}\cdot y\right)}{y^6}$

$\large \displaystyle \lim_{y \to 0} \frac{y^6\left(\frac{1}{120}-\frac{1}{36}+\frac{3}{40}\right)}{y^6}$

$\frac{1}{120}-\frac{1}{36}+\frac{3}{40} = \boxed{\frac{1}{18}}$